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ioda
2 years ago
14

Solve using elimination ​

Mathematics
1 answer:
oksian1 [2.3K]2 years ago
5 0
When you’re using elimination you eliminate using like terms so the -2y and +2y are set up to cancel each other out already, and after that since the y’s have been eliminated you can group together the rest of the like terms such as the X’s and non variable numbers, giving you 6x=-18 and after you simplify you end out with X=-3
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Evaluate f(x)= -6+12 when x= -3,x=0, and x=1.
Anton [14]

Given :

A function, f(x) = -6 + 12

To Find :

The value of f(x) when x= -3, x= 0 and x=1.

Solution :

Given function is f(x) = -6 + 12 .

Simplifying above function, we get :

f(x) = 6

Now, the given function is independent of x.

So, for any value of x the the value of functions remains constant.

Hence, this is the required solution.

5 0
3 years ago
F(X) = (x + 5)(x - 1)
marusya05 [52]

Answer:

X=-5 or X=1

Step-by-step explanation:

f(x)=(x+5)(x-1)

(x+5)(x-1)=0

x=-5 orx=1

4 0
2 years ago
True or false if you divide a number by two it's always even
Gnesinka [82]
False, ex: 18/2 = 9
false, ex: 18/2 = 9
8 0
3 years ago
Read 2 more answers
Helppppppppppp. Pleaseeeeeeeeeeeeeeeeeeeeeeeeeeee
olga55 [171]

Answer:

7

Step-by-step explanation:

form a proportion

12/2=42/x

cross multiply

12x=84

x=7

6 0
3 years ago
Read 2 more answers
How many equivalence relations are there on the set 1, 2, 3]?
Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0
3 years ago
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